Question: $f(x, y) = (4yx^2, x + 3xy^2)$ What is the curl of $f$ at $(-2, 3)$ ?
The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ x + 3xy^2 \right] \\ \\ &= 1 + 3y^2 \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ 4yx^2 \right] \\ \\ &= 4x^2 \end{aligned}$ Therefore: $\text{curl}(f) = 1 + 3y^2 - 4x^2$ The curl of $f$ at $(-2, 3)$ is $12$.